汽车理论matlab作业

一、确定一轻型货车的动力性能。 1)绘制汽车驱动力与行驶阻力平衡图; 2)求汽车最高车速与最大爬坡度;

3)绘制汽车行驶加速度倒数曲线;用计算机求汽车用Ⅱ档起步加速行驶至 70km/h 所需 的加速时间。

已知数据略。(参见《汽车理论》习题第一章第3题) 解题程序如下:用Matlab语言

(1)绘制汽车驱动力与行驶阻力平衡图

m1=2000; m2=1800; mz=3880;

g=9.81; r=0.367; CdA=2.77; f=0.013; nT=0.85; ig=[5.56 2.769 1.644 1.00 0.793]; i0=5.83;

If=0.218; Iw1=1.798; Iw2=3.598; Iw=2*Iw1+4*Iw2; for i=1:69

n(i)=(i+11)*50;

Ttq(i)=-19.313+295.27*(n(i)/1000)-165.44*(n(i)/1000)^2+40.874*(n(i)/1000)^3-3.8445*(n(i)/1000)^4;

end for j=1:5 for i=1:69

Ft(i,j)=Ttq(i)*ig(j)*i0*nT/r; ua(i,j)=0.377*r*n(i)/(ig(j)*i0);

Fz(i,j)=CdA*ua(i,j)^2/21.15+mz*g*f; end

end plot(ua,Ft,ua,Ff,ua,Ff+Fw)

title('汽车驱动力与行驶阻力平衡图'); xlabel('ua(km/h)'); ylabel('Ft(N)'); gtext('Ft1')

gtext('Ft2') gtext('Ft3') gtext('Ft4') gtext('Ft5') gtext('Ff+Fw')

(2)求最大速度和最大爬坡度 for k=1:175

n1(k)=3300+k*0.1;

Ttq(k)=-19.313+295.27*(n1(k)/1000)-165.44*(n1(k)/1000)^2 +40.874*(n1(k)/1000)^33.8445*(n1(k)/1000)^4;

Ft(k)=Ttq(k)*ig(5)*i0*nT/r; ua(k)=0.377*r*n1(k)/(ig(5)*i0);

Fz(k)=CdA*ua(k)^2/21.15+mz*g*f; E(k)=abs((Ft(k)-Fz(k))); end for k=1:175 if(E(k)==min(E))

disp('汽车最高车速='); disp(ua(k)); disp('km/h');

end end

for p=1:150 n2(p)=2000+p*0.5;

Ttq(p)=-19.313+295.27*(n2(p)/1000)-165.44*(n2(p)/1000)^2+40.874*(n2(p)/1000)

^3-3.8445*(n2(p)/1000)^4; Ft(p)=Ttq(p)*ig(1)*i0*nT/r;

ua(p)=0.377*r*n2(p)/(ig(1)*i0); Fz(p)=CdA*ua(p)^2/21.15+mz*g*f; af(p)=asin((Ft(p)-Fz(p))/(mz*g)); end for p=1:150

if(af(p)==max(af)) i=tan(af(p));

disp('汽车最大爬坡度='); disp(i); end end

汽车最高车速=99.0679km/h

汽车最大爬坡度=0.3518

(3)计算2档起步加速到70km/h所需时间

for i=1:69 n(i)=(i+11)*50;

Ttq(i)=-19.313+295.27*(n(i)/1000)-165.44*(n(i)/1000)^2+40.874*(n(i)/1000)^3-3.8445*(n(i)/1000)^4; end

for j=1:5 for i=1:69

deta=1+Iw/(mz*r^2)+If*ig(j)^2*i0^2*nT/(mz*r^2); ua(i,j)=0.377*r*n(i)/(ig(j)*i0);

a(i,j)=(Ttq(i)*ig(j)*i0*nT/r-CdA*ua(i,j)^2/21.15 -mz*g*f)/(deta*mz); if(a(i,j)<=0)

a(i,j)=a(i-1,j); end

if(a(i,j)>0.05) b1(i,j)=a(i,j); u1(i,j)=ua(i,j); else

b1(i,j)=a(i-1,j); u1(i,j)=ua(i-1,j); end

b(i,j)=1/b1(i,j); end end

x1=u1(:,1);y1=b(:,1); x2=u1(:,2);y2=b(:,2); x3=u1(:,3);y3=b(:,3); x4=u1(:,4);y4=b(:,4); x5=u1(:,5);y5=b(:,5);

plot(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5); title('加速度倒数时间曲线'); axis([0 120 0 30]); xlabel('ua(km/h)'); ylabel('1/aj');

gtext('1/a1') gtext('1/a2') gtext('1/a3') gtext('1/a4') gtext('1/a5')

for i=1:69 A=ua(i,3)-ua(69,2); if (A<1&A>0) j=i; end

B=ua(i,4)-ua(69,3);

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